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3 years ago

Please answer This Just Try Don't skip

Mathematics

2 answers:

Sav [38]3 years ago

7 0

(i)
Area of rectangle = length * breadth
So,
x^2 - 12x + 35 = (x - 5) * breadth

Factorize area and divide by length to simplify.
=> {(x-5)(x-7)} / (x-5) = breadth
=> x-7 = breadth

Therefore, breadth = 7

(ii)
Length = 10 - 5 = 5m
Breadth = 10 - 7 = 3m
Area = 10^2 - 12(10) + 35 = 15sq.m

attashe74 [19]3 years ago

4 0

Answer:

5 m , 3 m , 15 m²

Step-by-step explanation:

Given area (A) = x² - 12x + 35 and A = length × breadth , then

x² - 12x + 35 = (x - 5)(x - 7) ← in factored form

We know length = x - 5 , then breadth = x - 7

When x = 10

length = x - 5 = 10 - 5 = 5 m and breadth = x - 7 = 10 - 7 = 3 m

A = 10² - 12(10) + 35 = 100 - 120 + 35 = 15 m²

A = length × breadth = 5 × 3 = 15 m²

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Consider the following equations. f(x) = − 4/ x3, y = 0, x = −2, x = −1. Sketch the region bounded by the graphs of the equation

Sindrei [870]

Answer:

1.5 unit^2

Step-by-step explanation:

Solution:-

- A graphing utility was used to plot the following equations:

$f ( x ) = - \frac{4}{x^3}\\\\y = 0 , x = -1 , x = -2$

- The plot is given in the document attached.

- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).

- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.

We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).

The double integration formulation can be written as:

$A= \int\limits_c^d \int\limits_a^b {} \, dy.dx \\\\A = \int\limits_c^d { - \frac{4}{x^3} } . dx\\\\A = \frac{2}{x^2} |\limits_-_2^-^1\\\\A = \frac{2}{1} - \frac{2}{4} \\\\A = \frac{3}{2} unit^2$

Answer: 1.5 unit^2 is the amount of area bounded by the given curve f ( x ) and the boundary equations.

Download docx

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3 years ago

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Nataly_w [17]

Answer:

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2 years ago

Subtract -2x^2+4x-1−2x

vivado [14]

Answer:

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Step-by-step explanation:

The question is not well written. I guess what you mean is:

Subtract -2x^2+4x-1 ( minus 2x squared plus 4x minus 1) from 6x^2+3x-9 (6x squared plus 3x minus 9).

To subtract -2x^2+4x-1 from 6x^2+3x-9, that is

6x^2+3x-9 - (-2x^2+4x-1). This can be written as

6x²+3x-9 - (-2x²+4x-1)

Now, opening the bracket, we will get

6x²+3x-9 +2x²-4x+1

Then, collecting like terms, we get

6x²+2x²+3x-4x-9+1

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The answer is 8x²-x -8.

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3 years ago

Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0

andreev551 [17]

Recall the angle sum identity for cosine:

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

cos(x - y) = cos(x) cos(y) + sin(x) sin(y)

==> sin(x) sin(y) = 1/2 (cos(x - y) - cos(x + y))

Then rewrite the equation as

sin(4x) sin(5x) + sin(4x) sin(3x) - sin(2x) sin(x) = 0

1/2 (cos(-x) - cos(9x)) + 1/2 (cos(x) - cos(7x)) - 1/2 (cos(x) - cos(3x)) = 0

1/2 (cos(9x) - cos(x)) + 1/2 (cos(7x) - cos(3x)) = 0

sin(5x) sin(-4x) + sin(5x) sin(-2x) = 0

-sin(5x) (sin(4x) + sin(2x)) = 0

sin(5x) (sin(4x) + sin(2x)) = 0

Recall the double angle identity for sine:

sin(2x) = 2 sin(x) cos(x)

Rewrite the equation again as

sin(5x) (2 sin(2x) cos(2x) + sin(2x)) = 0

sin(5x) sin(2x) (2 cos(2x) + 1) = 0

sin(5x) = 0 or sin(2x) = 0 or 2 cos(2x) + 1 = 0

sin(5x) = 0 or sin(2x) = 0 or cos(2x) = -1/2

sin(5x) = 0 ==> 5x = arcsin(0) + 2nπ or 5x = arcsin(0) + π + 2nπ

… … … … … ==> 5x = 2nπ or 5x = (2n + 1)π

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sin(2x) = 0 ==> 2x = arcsin(0) + 2nπ or 2x = arcsin(0) + π + 2nπ

… … … … … ==> 2x = 2nπ or 2x = (2n + 1)π

… … … … … ==> x = nπ or x = (2n + 1)π/2

cos(2x) = -1/2 ==> 2x = arccos(-1/2) + 2nπ or 2x = -arccos(-1/2) + 2nπ

… … … … … … ==> 2x = 2π/3 + 2nπ or 2x = -2π/3 + 2nπ

… … … … … … ==> x = π/3 + nπ or x = -π/3 + nπ

(where n is any integer)

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3 years ago

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Oxana [17]

Answer:

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2 years ago

Please answer This Just Try Don't skip​

Please answer This Just Try Don't skip